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My understanding of imaginary numbers is less than perfect... but if I'm not mistaken...

x = 10i

or

x = 10/i

???

EDIT: no, that's not right. Gargle.

x = 10i

or

x = 10/i

???

EDIT: no, that's not right. Gargle.

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@Disch this question does not delve into imaginary numbers, as far as I know

@L B Excuse my ignorance, but what does the graph tell me?

@L B Excuse my ignorance, but what does the graph tell me?

x - 1 = (-3)^2

x - 1 = 9

x = 10?

Ignore all of that.

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((x-1)^0.5) + 3 = 0 iff sqrt(x-1) = -3 iff sqrt(y) = -3.

By definition of sqrt: for all real values of q, sqrt(q) is non-negative.

Therefore, there exists no real x that meets the condition.

By definition of sqrt: for all real values of q, sqrt(q) is non-negative.

Therefore, there exists no real x that meets the condition.

...which is why I thought x had to be imaginary. But apparently there's not even an imaginary solution?

Imaginary numbers are when you have a square root of a negative number, not when the square root of a number equals a negative number.

sqrt(-4) = 2i

sqrt(-4) = 2i

By definition of sqrt: for all real values of q, sqrt(q) is non-negative. |

Otherwise f(x) = x^0.5 wouldn't be a function.

I read an interesting thing here: http://www.xamuel.com/plus-or-minus-square-roots/

And was wondering why? (The later explanation fails to convince me)

For example, 4 has two square roots, 2 and -2. This is because 2²=4 and (-2)²=4. But the square root of 4 is 2 (never -2). |

And was wondering why? (The later explanation fails to convince me)

I was taught that sqrt(9) was equal to +/- (read: plus minus) 3 |

1. If x>=0, sqrt(x) is a non-negative number such that sqrt(x)^2 = x.

2. If x<0, sqrt(x) is undefined.

You could define a multifunction f:

Imaginary numbers are when you have a square root of a negative number, not when the square root of a number equals a negative number. |

EDIT:

And was wondering why? (The later explanation fails to convince me) |

* sqrt(x)=3

* sqrt(x)=-3

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This is weird. Such an elementary mathematics topic, and I'd never really thought of this. I understand a function can only have one solution per input. But then how does a square root of a positive number have 2 solutions? Why does the square root have to be a function?

@helios I think I get it, the reason my first mentioned equation is false, is simply due to my prior lack of understanding of the definition of the sqrt function. thank you :)

EDIT: which is sqrt(x)=b; b>=0

EDIT: which is sqrt(x)=b; b>=0

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I'm just curious ResidentBiscuit, I've seen a lot of your posts suggesting you're in your 3rd/4th (?) year of university for computer science. Does your school not require you to take many math courses? Or is sqrt just something you never really thought about and always just assumed that the positive value was the "right" one? (This is not meant to be offensive, I'm just curious)

+(4^{½}) = +(2)

-(4^{½}) = -(2)

±(4^{½}) = ±(2)

∓(4^{½}) = ∓(2)

Think about it, what if +4^{½} = ±2?

2^{2} = 4 = 2 * 4^{½} = 2 * ±2

4 = ±4 -> 4 = 4 and**4 = -4**

An even root does not imply ± nor ∓, you have to specify that with the root.

-(4

±(4

∓(4

Think about it, what if +4

2

4 = ±4 -> 4 = 4 and

An even root does not imply ± nor ∓, you have to specify that with the root.

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Hi,

((x-1)^0.5)+3=0

=> ((x-1)^0.5) = -3

=> (x-1) = 9

=> x=10

of course for sqrt(10-1) then we have + or -3, there's the solution taking the negative sqrt of (10 - 1) and adding 3.

Cheers,

((x-1)^0.5)+3=0

=> ((x-1)^0.5) = -3

=> (x-1) = 9

=> x=10

of course for sqrt(10-1) then we have + or -3, there's the solution taking the negative sqrt of (10 - 1) and adding 3.

Cheers,

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